Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, dp[100005][21], dep[100005];
vector<int> G[100005];
vector<int> dp1[100005][21];
vector<int> ans;
vector<int> c[100005];
vector<int> merge(vector<int> &b, vector<int> &c) {
  int k1 = 0, k2 = 0;
  vector<int> a;
  a.clear();
  while (k1 < b.size() && k2 < c.size() && a.size() < 10) {
    if (b[k1] < c[k2]) {
      a.push_back(b[k1]);
      k1++;
    } else {
      a.push_back(c[k2]);
      k2++;
    }
  }
  while (a.size() < 10 && k1 < b.size()) {
    a.push_back(b[k1]);
    k1++;
  }
  while (a.size() < 10 && k2 < c.size()) {
    a.push_back(c[k2]);
    k2++;
  }
  return a;
}
void dfs(int u, int fa) {
  dp1[u][0] = c[u];
  dp1[u][1] = merge(c[u], c[fa]);
  dep[u] = dep[fa] + 1;
  dp[u][0] = fa;
  dp[u][1] = dp[fa][0];
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    dfs(v, u);
  }
}
void build(int root) {
  dfs(root, 0);
  for (int j = 2; j <= 20; j++)
    for (int i = 1; i <= n; i++) {
      dp[i][j] = dp[dp[i][j - 1]][j - 1];
      dp1[i][j] = merge(dp1[i][j - 1], dp1[dp[i][j - 1]][j - 1]);
    }
}
void mov(int &u, int v) {
  for (int i = 20; i >= 0; i--)
    if (dep[dp[u][i]] >= dep[v]) {
      ans = merge(ans, dp1[u][i]);
      u = dp[u][i];
    }
}
int solve(int u, int v) {
  if (dep[u] > dep[v]) mov(u, v);
  if (dep[u] < dep[v]) mov(v, u);
  if (u == v) {
    ans = merge(ans, dp1[v][0]);
    return u;
  }
  for (int i = 20; i >= 0; i--) {
    if (dp[u][i] != dp[v][i]) {
      ans = merge(ans, dp1[u][i]);
      ans = merge(ans, dp1[v][i]);
      u = dp[u][i];
      v = dp[v][i];
    }
  }
  ans = merge(ans, dp1[u][0]);
  if (u != v) ans = merge(ans, dp1[v][1]);
  return dp[u][0];
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    c[x].push_back(i);
  }
  build(1);
  while (q--) {
    int u, v, a;
    ans.clear();
    scanf("%d%d%d", &u, &v, &a);
    solve(u, v);
    int len = ans.size();
    len = min(len, a);
    printf("%d", len);
    for (int i = 0; i < len; i++) printf(" %d", ans[i]);
    printf("\n");
  }
}